Respuesta :
[tex]\vec r(t)=t^2\,\vec\imath+t^3\,\vec\jmath-2t\,\vec k[/tex]
[tex]\mathrm d\vec r=(2t\,\vec\imath+3t^2\,\vec\jmath-2\,\vec k)\,\mathrm dt[/tex]
[tex]\vec f(x,y,z)=(x+y^2)\,\vec\imath+xz\,\vec\jmath+(y+z)\,\vec k[/tex]
[tex]\vec f(x(t),y(t),z(t))=(t^2+t^6)\,\vec\imath-2t^3\,\vec\jmath+(t^3-2t)\,\vec k[/tex]
The line integral is
[tex]\displaystyle\int_C\vec f\cdot\mathrm d\vec r=\int_0^2\bigg((t^2+t^6)\,\vec\imath-2t^3\,\vec\jmath+(t^3-2t)\,\vec k\bigg)\cdot(2t\,\vec\imath+3t^2\,\vec\jmath-2\,\vec k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^2(2t^7-6t^5+4t)\,\mathrm dt=\boxed{8}[/tex]
Given:
[tex]\to \bold{\int_{c} \ F \cdot dx}[/tex]
Where
[tex]\to \bold{F(x,y,z)=(x+y^2)i+xzj+(y+z)K}[/tex] and C is given by the vector function [tex]\bold{r(t)}[/tex].
[tex]\to \bold{r(t)=t^2i+t^3j-2tk,\ \ 0\leq t\leq 2}\\\\\to \bold{dx=(2ti+3t^2j-2k)dt}\\\\[/tex]
therefore
[tex]\to \bold{x=t^2}\\\\ \to \bold{y=t^3}\\\\ \to \bold{z=2t}\\\\\to \bold{F(r)=(t^2+(t^3)^2)i+t^2(-2t)j+(t^3-2t)k}\\\\[/tex]
[tex]\bold{=(t^2+t^6)i-2t^3j+(t^3-2t)k}\\\\[/tex]
[tex]\bold{\therefore}\\\\\bold{F(r).dx=((t^2+t^6)i-2t^3j+(t^3-2t)k).(2ti+3t^2j-2k)}\\\\[/tex]
[tex]=\bold{((t^2+t^6)2t-2t^3(3t^2)+(t^3-2t)(-2)]dt}\\\\=\bold{(2t^3+2t^7-6t^5-2t^3+4t)dt}\\\\=\bold{(2t^7-6t^5+4t)dt}\\\\[/tex]
Calculating the line integral:
[tex]\bold{\int_{c}\ F .dx=\int^{2}_{t=0} (2t^7-6t^5+4t)dt}\\\\[/tex]
[tex]=\bold{[\frac{t^8}{4}-t^6+2t^2]^{2}_{0}}\\\\=\bold{\frac{2^8}{4}-2^6+2(2)^2}\\\\=\bold{\frac{256}{4}-64+2(4)}\\\\=\bold{64-64+8}\\\\=\bold{8}[/tex]
Therefore the final answer "8".
Learn more:
line integral: brainly.com/question/16571684
